61. Rotate List (M)

Rotate List (M)

Given a linked list, rotate the list to the right by k places, where k is non-negative.

Example 1:

Input: 1->2->3->4->5->NULL, k = 2
Output: 4->5->1->2->3->NULL
Explanation:
rotate 1 steps to the right: 5->1->2->3->4->NULL
rotate 2 steps to the right: 4->5->1->2->3->NULL

Example 2:

Input: 0->1->2->NULL, k = 4
Output: 2->0->1->NULL
Explanation:
rotate 1 steps to the right: 2->0->1->NULL
rotate 2 steps to the right: 1->2->0->NULL
rotate 3 steps to the right: 0->1->2->NULL
rotate 4 steps to the right: 2->0->1->NULL

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题意

依次将链表最后一个元素移到最前面，共移动k次。

思路

由于k可能大于链表本身长度length，所以先遍历一遍链表得到length，简化 k = k % length，再找到倒数k个元素前的一个元素(即新链表的最后一个元素)，以此为分界点，将前半链表接到后半链表的后面。

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代码实现

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */

class Solution {
    public ListNode rotateRight(ListNode head, int k) {
        if (k == 0 || head == null) {
            return head;
        }
        
        int length = 1;
        ListNode last = head;
        while (last.next != null) {
            length++;
            last = last.next;
        }
        k = k % length;
        if (k == 0) {
            return head;
        }
        
        // 找到新链表的尾结点
        int count = 1;
        ListNode newLast = head;
        while (count < length - k) {
            newLast = newLast.next;
            count++;
        }
        
        // 找到新链表的头结点，并将前半链表移到最后
        ListNode newHead = newLast.next;
        newLast.next = null;
        last.next = head;
        
        return newHead;
    }
}